Lyapunov exponents of convergent sequence of matrices

Question

Let $A_n$ be a sequence of $d \times d$ matrices converging to a matrix $A$, all invertible and diagonalizable. We can define the Lyapunov spectrum of the corresponding dynamical system: $$ \chi = \big\{ \lim_{n \to \infty} \frac1n \log\|A^{(n)}v\| :\ v \in \Bbb R^d \big\}$$

where $A^{(n)} = A_n A_{n-1} A_{n-2} \cdots A_1$. Is it the case, in general, that $\chi$ equals the Lyapunov spectrum of the matrix $A$? If not, can we make this true by assuming that the sequence converges fast enough (for some notion of "fast enough")?

Answer 1

So the answer is yes. A brief version is that it suffices to check this for the top exponent and then to use exterior powers to deduce the result for subsequent exponents.

(Slightly modified to make $\lambda$ the leading Lyapunov exponent)

Next, if $E$ is the sum of the generalized eigenspaces of all the eigenvalues of maximal modulus, and $F$ is the sum of the remaining generalized eigenspaces, for any $\epsilon>0$, there is a power of $A$ such that $e^{n(\lambda-\epsilon)}\|e\|\le \|A^ne\|\le e^{n(\lambda+\epsilon)}\|e\|$ for all $e\in E$ and $\|A^nf\|\le e^{n\mu}\|f\|$ where $\lambda$ is the logarithm of the absolute value of the dominant eigenvalue(s) of $A$ and $\mu<\lambda$. Now you can build a “cone” of points whose $F$ component is at most $\epsilon$ their $E$ component. This cone is invariant under small perturbations of $A^n$. Now a bit of triangle inequality gives you the result you’re looking for.